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Innovation and Digital Transformation for a Sustainable World




           problem (24) can be represented as                 Algorithm 1 The Gaussian Randomization Method For
                                                              Solving (41).
                                                 !
                                    ¯ v Φ   ,   ¯ v                                          ∗
                                       
                        ∑︁
                max    log 2                           (26)a   1: Initialize: The optimal solution ¯ v of the problem (41),
                 v          1 + Í     ¯ v Φ   ,   ¯ v +    2
                                        
                      =1           =  +1                         the random number Ω and    = 1, 2, · · · , Ω.
                                                               2: repeat
                                             !
                                     ∑︁      2                 3:   Generate random vectors q ∼ CN (0, I   +1 ).
            s.t.¯ v Φ   ,   ¯ v ≥     ¯ v Φ   ,   ¯ v +     , ∀   ∈ K (26)b
                                              
                                =  +1                          4:   Obtain random values for the solution of optimization
                                                                                 ∗
                                                                 problem (41) ¯ v    = ¯ v · q.
                            |      | = 1, ∀   ∈ N.     (26)c
                                                               5:   Standardize ¯ v    to obtain ¯ v           .
           Then, we define V = ¯ v¯ v . By utilizing the property of                       
                                  
                                                               6:   Transform ¯ v     into   ×   dimensional matrix Θ    =
                         
           matrix traces ¯ v Φ   ,   ¯ v =      Φ   ,   ¯ v¯ v     and SDR to forcibly  norm
                                                                 diag ¯ v t  N×N  matrix.
           omit rank-one constraint [25], problem (26) can be relaxed as
                                                               7:    Compute total computational bits                 according
                                                                                                    
                                                  !              to Θ    .

                                        Φ   ,   V
                       ∑︁
               max    log 2                            (27)a   8:       ←    + 1.
                 V         1 + Í          Φ   ,   V +    2
                     =1           =  +1                        9: until    = Ω.
                                                              10: Output:  index    which maximizes                and its
                                                                                                     
                                ∑︁
                                            2                    corresponding Θ    .
             s.t.     Φ   ,   V ≥       Φ   ,   V +    , ∀   ∈ K  (27)b        ∗
                                              
                              =  +1
                                                                                   Í
                                                                                                − Φ   
                      V   ,   = 1, ∀   ∈ {1, · · · ,    + 1} .  (27)c       4  −       ∑︁    =  +1  Φ   ,   (  ,  )    ,   (  ,  )
                                                                           =                            ,  (31)
                                                                             ln2         Í            2
                                V ⪰ 0,                 (27)d       Im V   ,       =1      V    =  +1  Φ   ,   +      
           where V   ,   denotes the diagonal elements of matrix V.
                                                              where Φ   ,   (  ,  )  represents (  ,   ) −   ℎ element of Φ   ,   .
           Although the constraints (27)b-(27)d are all convex, the
                                                              Based on the above derivations, we can reformulate problem
           problem remains non-convex because of (27)a. Thus, by
                                                              (27) in   -th iteration as
           using the properties of matrix traces [25], we can equivalently
           transform (27)a into                                 max      3 (V) −    4 (V (  −1) )
                                                                  V
                                                  !

                                        Φ   ,   V                          ∑︁   −1
                       ∑︁
                                                                           ∑︁
                      log                                                                (  −1)          4
                         2  1 + Í                2                    −       Re V   ,   − V   ,    ×
                      =1          =  +1      Φ   ,   V +                                              (  −1)
                                                   
                                                                          =1   =1                 Re V   ,  
                                      ∑︁  !   !
                         ∑︁
                    =   log 2       V  Φ   ,   +    2     (28)             ∑︁   −1       (  −1)          4
                                                                           ∑︁
                        =1           =                                −       Im V   ,   − V   ,    ×     (  −1)
                                                                          =1   =1
                                       ∑︁  !   !                                                  Im V   ,  
                         ∑︁
                    −   log       V    Φ   ,   +     2  .                                                 (32)a
                           2                    
                       =1           =  +1                                      s.t.(27)   − (27)  .       (32)b
           And we define                                      In this case, optimization problem (32) becomes a SDP
                                                              problem. Thus, we can efficiently solved by applying convex
                                          !     !
                           ∑︁
                                        ∑︁
                   3 (V) =  log 2       V  Φ   ,   +    2     ,  optimization tools, e.g., interior point methods or the CVX
                         =1           =                       package [24]. However, since the obtained optimal solution
                                                        (29)
                                                              V is not necessarily rank-one, we need to further apply
                                            !    !              ∗
                           ∑︁
                                        ∑︁
                   4 (V) =  log       V  Φ   ,   +    2  .    Gaussian randomization to obtain Θ. The detailed steps of
                             2                    
                         =1           =  +1                   Gaussian randomization are outlined in Algorithm 1.
           We can easily obtain that (28) is a DC function, so we can use  In summary, this section proposes an AO-based algorithm
           DC programming to solve it. Nevertheless, we cannot take  to maximize computation rate.  Firstly, we derive the
           partial derivatives with respect to the complex variables as in  closed-form solution for W.  Afterwards we achieve
           (20) by reason of the complex matrix of V. Moreover, since  the closed-form optimal solution of       by solving the
              4 (V) is not an analytic function, its derivative with respect to  subproblem (11). Then, we optimize p by using variable
           V does not exist [26]. To address this issue, we derive    4 (V)  substitution and SCA-based iterative algorithm. Finally, Θ is
           with regard to the real and imaginary parts of V. Since V  optimized by applying variable substitution and SDR-based
           is a symmetric matrix, we only need to calculate the real  iterative algorithm, where the four subproblems are optimized
                                                              alternately.
           and imaginary parts of the lower triangular elements of V.
           Specifically, for      ,   ,    ≤   , ∀   ∈ N, the partial derivatives
           concerning the real and imaginary parts are respectively given  3.5 Complexity Analysis
           by
                                                              We define the computational complexity of each iteration as
                                Í                                      
                                             + Φ                  , and the maximum number of iterations as   . Then, the
                      4   1      ∑︁    =  +1  Φ   ,   (  ,  )    ,   (  ,  )
                        =                            ,  (30)  computational complexity upper bound can be denoted by
                          ln2         Í            2
                               =1      V    =  +1  Φ   ,   +        O(      ). Note that the computational complexity comes
                Re V   ,                                                  
                                                          – 283 –
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